Perfect numbers beside 6 in mod6.

i chekced a few perfect numbers with module 6 and, a nice property is that all mod 6 equal 4 (at least for those i checked), i guees that if an odd perfect number would exist then its mod 6 would be different.

i wonder how to prove that for every even perfect number greater than 6, its mod 6 equal 4?

i guess because its even it's divisble by 2, and then the question becomes how to prove that mod 3 equal 2, then how do you prove/disprove the assertion?

muzza, why this "2^(n - 1) * (2^n - 1) = (-1)^(n - 1) * ((-1)^n - 1)"?
or you were reffering to conguerence here, and even if you did refer to conguernece shouldn't it be mod2
because:
2^(2n-1)-2^(n-1)-(-1)^(2n-1)-(-1)^n=2^(2n-1)-2^(n-1)+(-1)^2n+(-1)^(n+1)=2^(2n-1)-2^(n-1)+2=0mod2

This shows that 2^n must alternate between the above two forms. In other words, modulo 6, 2^n must alternate between 4 and -4. Since [itex]2^1 \equiv -4~(mod~6) [/itex], we have [itex]2^n \equiv (-1)^n\cdot 4~(mod~6) [/itex] and hence [itex](-1)^n\cdot 2^n \equiv 4 ~(mod~6) [/itex]